Question

The amount of mineral water consumed by a person per day on the job is normally distributedwith mean 19 ounces and standard deviation 5 ounces. A company supplies its employees with 2000 ounces ofmineral water daily. The company has 100 employees. Find the probability that the mineral water suppliedby the company will not satisfy the water demanded by its employees.

Answer 1

Answer:

0.0228 = 2.28% probability that the mineral water supplied by the company will not satisfy the water demanded by its employees.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

n values from a normal distribution:

The mean is $\mu*n$ and the standard deviation is $\sigma\sqrt{n}$

Normally distributed with mean 19 ounces and standard deviation 5 ounces.

This means that $\mu = 19, \sigma = 5$

The company has 100 employees.

This means that for the mean consumption of all employees, we have that:

$\mu = 19*100 = 1900$

$\sigma = 5\sqrt{100} = 50$

Find the probability that the mineral water supplied by the company will not satisfy the water demanded by its employees.

Consumption higher than 2000 ounces, which is 1 subtracted by the pvalue of Z when X = 2000. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{2000 - 1900}{50}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

1 - 0.9772 = 0.0228

0.0228 = 2.28% probability that the mineral water supplied by the company will not satisfy the water demanded by its employees.